Data: Circle with centre ‘O’ is circumscribed in a quadrilateral ABCD.
To Prove: ∠AOD + ∠BOC = 180°
Sides of the quadrilateral AB, BC, CD and DA touches at the points P, Q, R and S respectively.
OA. OB, OC, OD and OP, OQ, OR, OS are joined.
OA bisects ∠POS.
∠1 =∠2
∠3 = ∠4
∠5 = ∠6
∠7 = ∠8
2(∠1 + ∠4 + ∠5 + ∠8) = 360°
(∠1 + ∠8) (∠4 + ∠5) = 180°
∴ ∠AOD = ∠BOD = 180°
∴ Opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.